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  4. CaCO 3 is decomposed by HCl (density 1.825 g / cc ) CaCO 3+2 HCl arrow CaCl 2+ H 2 O + CO 2 Volume of HCl required to decompose 10 g of 50 % pure CaCO 3 is

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Q. $CaCO _3$ is decomposed by $HCl$ (density $1.825\, g / cc$ )
$CaCO _3+2 HCl \rightarrow CaCl _2+ H _2 O + CO _2$
Volume of $HCl$ required to decompose $10 \,g$ of $50 \%$ pure $CaCO _3$ is

Some Basic Concepts of Chemistry

Solution:

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$(50\%$ pure) $5\, g \frac{7.3}{2} g \, HCl$
$\frac{\text { mass }}{\text { volume }}=$ density
$\therefore$ Volume of $HCl =\frac{\text { mass of } HCl }{\text { density of } HCl }=\frac{7.3}{2 \times 1.825}$
$=2 \, mL$